Question

At ${25}^{\circ }C$, for the process $H_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(g\right)$; $\Delta G^{\circ }$ is $8.6kJ$. The vapour pressure of water at this temperature is nearly:

• A. $29$ torr
• B. $285$ torr
• C. $23.17$ torr
• D. $100$ torr

Answer 1

Answer: (C)

$23.17$ torr

Consider the vaporization as equilibrium reaction.

The the activities satisfy equilibrium equation

$K=\frac{a\left(H_{2}O\right(g)}{a\left(H₂O\right(l)}$

The vapour pressure is the partial pressure over gaseous water over pure water.

Since liquid water water is pure , it has activity 1.

$a\left(H_{2}O\right(l\left)\right)=1$

Assuming ideal gas phase, the activity of gaseous water is equal to ratio of partial pressure to total pressure:

$a\left(H_{2}O\right(g\left)\right)=\frac{p\left(H_{2}O\right(g\left)\right)}{p^o}$

Hence

$K=\frac{p\left(H_O\right(g\left)\right)}{p^o}$

$=p\left(H_{2}O\right(g\left)\right)=p^o\cdot K$

Equilibrium constant and $\Delta G$ are related as

$\Delta G=-R\cdot T\cdot ln\left(K\right)$

$K=e^{\frac{-\Delta G}{R\cdot T}}$

Therefore

$p\left(H_{2}O\right(g\left)\right)=p^o\cdot e^{\frac{-\Delta G}{R\cdot T}}$

$=101325Pa\cdot e^{\frac{-8600J/mol}{(8.314472J/molK\cdot 298.15K)}}$

$=3155Pa$

converting Pa to torr, we get 23.17 as answer

Hence, te correct option is $\text{C}$